characterization of field

Proposition 1.

Let $\mathcal{R}\neq 0$ be a commutative ring with identity. The ring $\mathcal{R}$ (as above) is a field if and only if $\mathcal{R}$ has exactly two ideals: $(0),\mathcal{R}$ .

Proof.

( $\Rightarrow$ ) Suppose $\mathcal{R}$ is a field and let $\mathcal{A}$ be a non-zero ideal of $\mathcal{R}$ . Then there exists $r\in\mathcal{A}\subseteq\mathcal{R}$ with $r\neq 0$ . Since $\mathcal{R}$ is a field and $r$ is a non-zero element, there exists $s\in\mathcal{R}$ such that

$s\cdot r=1\in\mathcal{R}$

Moreover, $\mathcal{A}$ is an ideal, $r\in\mathcal{A},s\in\mathcal{S}$ , so $s\cdot r=1\in\mathcal{A}$ . Hence $\mathcal{A}=\mathcal{R}$ . We have proved that the only ideals of $\mathcal{R}$ are $(0)$ and $\mathcal{R}$ as desired.

( $\Leftarrow$ ) Suppose the ring $\mathcal{R}$ has only two ideals, namely $(0),\mathcal{R}$ . Let $a\in\mathcal{R}$ be a non-zero element; we would like to prove the existence of a multiplicative inverse for $a$ in $\mathcal{R}$ . Define the following set:

$\mathcal{A}=(a)=\{r\in\mathcal{R}\mid r=s\cdot a,\text{ for some }s\in\mathcal{R}\}$

This is clearly an ideal, the ideal generated by the element $a$ . Moreover, this ideal is not the zero ideal because $a\in\mathcal{A}$ and $a$ was assumed to be non-zero. Thus, since there are only two ideals, we conclude $\mathcal{A}=\mathcal{R}$ . Therefore $1\in\mathcal{A}=\mathcal{R}$ so there exists an element $s\in\mathcal{R}$ such that

$s\cdot a=1\in\mathcal{R}$

Hence for all non-zero $a\in\mathcal{R}$ , $a$ has a multiplicative inverse in $\mathcal{R}$ , so $\mathcal{R}$ is, in fact, a field. ∎

Title	characterization of field
Canonical name	CharacterizationOfField
Date of creation	2013-03-22 13:57:03
Last modified on	2013-03-22 13:57:03
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	7
Author	alozano (2414)
Entry type	Theorem
Classification	msc 12E99
Synonym	a field only has two ideals
Related topic	Field
Related topic	Ring
Related topic	Ideal